HYDROLYSIS OF ORGANIC HALOGEN COMPOUNDS

Extracts from this document...

Introduction

HYDROLYSIS OF ORGANIC HALOGEN COMPOUNDS Name: Ho Ka Wing (9) Group:3 Grades: Date: 07-09-10 Objective The purpose of this experiment is to find out how the rate of hydrolysis of an organic halogen compound depends on: a the identity of the halogen atom. b the nature of the carbon-hydrogen 'skeleton'. Theory Experiment: In this experiment, the rate of hydrolysis of 1-chlorobutanne, 1-bomobutane, 1-iodobutane, Chlorobenzene is compared. A general equation for the hydrolysis is: R-X + H2O ???? R-OH + H+ + X- (Where R = alkyl or aryl group; X= halogen atom) Hydrolysis of organic halogen compounds is a nucleophilic substitution reaction. In a nucleophilic substitution a lone pair of electron on a nucleophile, H2Ois attracted to a carbon atom, with a partial positive charge. The nucleophile is then substituted for the atom or group attached to the carbon atom. Since halide ions are being substituted out. By following the rate of the reaction by carrying it out in the presence of silver ions, so that any halide ions produced form a silver halide precipitate. Ag+(aq) + X-(aq) ???? AgX(s) By comparing the time for appearance of precipitate, we can compare the rate of hydrolysis. The smaller the time needed, the faster the rate of hydrolysis. ...read more.

Middle

7. The tubes were continuously watched for about ten minutes and, in a copy of Results Table 1, the time was noted when a precipitate first appeared in each tube as a definite cloudiness. The water was heat to 60oC again at intervals. 8. Continue observation at intervals for about 30 minutes more, any further changes was noted in the appearance of the precipitates. Results Table Results Table 1 Reaction Time for precipitate to appear Observations A 1-chlorobutane 720 seconds White precipitate was formed B 1-bromobutane 12 seconds Pale yellow precipitate was formed C 1-iodobutane 3 seconds Yellow precipitate was formed D chlorobenzene N/A No obsevation Interpretation of the result bond Bond energy/kJ C-Cl C-Br C-I In halogenoalkanes 338 276 238 365 C-Cl In Comparison between haloalkane They all shows positive results but the rate of hydrolysis is different. Ag+ (aq) + Cl- (aq) � AgCl (s) White precipitate Ag+ (aq) + Br- (aq) � AgBr (s) White precipitate Ag+ (aq) + I- (aq) � AgI (s) Yellow precipitate The reactivity of halo group on the rate of hydrolysis reaction is: 1-iodobutane> 1-bromobutane>1-chlorobutane Bond energy: C-Cl > C-Br > C-I Bond strength: C-Cl > C-Br > C-I Reactivity: RCl < RBr < RI Since we can see that bond energy required to break the C-X bond decrease down the group. ...read more.

Conclusion

* Since the reactants is primary haloalkane which flavors Sn2. The primary carbocation which is less stable is formed. * The reactants are primary haloalkane so they are less bulky. So the steric hindrance is small, results in low energy of transition state and increase the activity in Sn2 mechanism. * Choice of nucleophile strong nucleophile in high concentration flavors Sn1 while weak nucleophile in dilute solution flavors Sn2. And this time in this experiment, H20 is weak nucleophile so it flavors Sn2. * Bimolecular Nucleophilic Substitution , SN2 SN2 applies mainly to methyl and primary haloalkanes. e.g.C4H9Br + OH????C4H9OH + Br- Mechanism: In this mechanism, the rate determining step involves 2 molecules (H2O and RX.).Hence, we use the term 'bimolecular'. And the rate of reaction should depend on the concentrations of both chlorobutane and hydroxide ion. Hence a second order reaction is observed. rate = k[H2O][RX] The energy change for this SN2 mechanism is shown below. Energy level diagram of the Sn2 reaction 2. Hydolysis for phenol( industrial process) halobenzene only undergo nucleophilic substitution of the halogen atom at extreme condition(conc.NaOH,350 oC, 200atm to give phenol) 3. Photodecomposition of AgBr At the end of experiment, there is some grey precipitate deposited in one of the tubes.It is mainly due to the fact that silver bromide is decomposed to give silver. 2AgBr(s) ????2Ag(s) + Br2 (g) Reference http://chemophile.pbworks.com/f/Halogenoalkanes.pdf http://en.wikipedia.org/wiki/Haloalkane ...read more.

Related AS and A Level Organic Chemistry essays

a balance correct to 3 decimal places, to ensure a high standard of accuracy. From this, the mass of ethanol could be calculated by subtracting the mass of the density bottle while empty from the mass of the density bottle with the ethanol solution.

Since 2 moles of NaOH reacts with 1 mole of Tartaric Acid, 8.535x10-3 moles of NaOH will react with 4.268x10-3 moles of Tartaric Acid. Therefore there are 4.268x10-3 moles of Tartaric Acid in 100cm3 of wine, so the concentration of acid in the wine can be calculated.

This is an alkaline hydrolysis known as sponification. It is the reaction of an ester with sodium hydroxide to produce an alcohol and the sodium salt of the carboxylic acid of the ester. The purity of the re-crystallised aspirin was 96.76% compared with the purity of pure aspirin which was 97.56%.

The concentration of halogenoalkanes should stay constant. In order words, the moles of halogenoalkanes used for this experiment have to be the same such that within a fixed volume of solution, the same number of moles of halogenoalkanes is found. This is to ensure that the chance of the halogenoalkanes molecules getting collide with the water molecules is the same for each set of experiment.

The longer the bond, the weaker the bond and this means less energy is required to break the long chlorine-iodine bond than the short carbon-bromine or even shorter carbon-chlorine bonds. Therefore I predict that iodine will be displaced in the least amount of time, followed by bromine and then chlorine.